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KSBGC
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It is observed that 2/7 f the balls that have red color also have 28 Jan 2018, 05:03
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

51% (03:11) correct 49% (02:58) wrong based on 244 sessions

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There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f the balls that have red color also have green color while 3/7 of the balls that have green color also have red color. What fraction of the balls in the jar has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT
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D
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doomedcat
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It is observed that 2/7 f the balls that have red color also have 03 Feb 2018, 22:32
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poul_249
Still not able to understand. Could someone explain ?

You have to visualise the problem as set problem...
Given 87 Balls with Red and Green colour
So the set equation becomes; Total (Red & Green Balls) = #Red Balls + # Green Balls - Both (Red & Green)

Given is 2/7 of Red balls also are Green ==> Both ( Red and green) = \(\frac{2}{7}\) * # Red Balls
Also it is stated that 3/5 of Green balls are red, which is again stating that ==> Both ( Red and green) = \(\frac{3}{7}\) * # Green Balls

Essentially it means that #Red Balls = \(\frac{7}{2}\)* Both (Red and Green)
Also it means that #Green Balls = \(\frac{7}{3}\)* Both (Red and Green)

Now coming back to original equation Total (Red & Green Balls) = #Red Balls + # Green Balls - Both (Red & Green)

Substituting the values we got above

87= \(\frac{7}{2}*Both + \frac{7}{3}*Both - Both\)
So Both = 18

We are asked to find Both/Total =\(\frac{18}{87}\) = \(6/29\)

Hope it helps , pls press Kudos
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It is observed that 2/7 f the balls that have red color also have Updated on: 03 Feb 2018, 22:34
Originally posted by doomedcat on 28 Jan 2018, 05:29.
Last edited by doomedcat on 03 Feb 2018, 22:34, edited 1 time in total.
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ven diagram..
87=R+G-Both RG

given that ...Both = 2R/7 = 3G/7; we need to find Both/total

87= 7Both/2 + 7Both/3 - Both

Both = 18

therefore , 18/87 = 6/29 D


Corrected G in 3G/7
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Still not able to understand. Could someone explain ?
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selim
There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f the balls that have red color also have green color while 3/7 of the balls that have green color also have red color. What fraction of the balls in the jar has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT


R Red
G Green
X Both

Given:
Total Balls = 87
(2/7)*R = X ----> R = (7/2)*X ---->(1)
(3/7)*G = X ----> G = (7/3)*X ---->(2)

formula : T = R+G - both(X)
Applying 1 and 2 in formula
87= (7/2)*X + (7/3)*X - X
Solving, we get X= 18

Now fr fraction X/T = 18/87 = 6/29

Answer : D 6/29
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It is observed that 2/7 f the balls that have red color also have 28 Aug 2019, 10:40
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KSBGC
There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f the balls that have red color also have green color while 3/7 of the balls that have green color also have red color. What fraction of the balls in the jar has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT


Bunuel's Solution:

The easiest way would be to spot that the correct answer must have the denominator which is a factor of 87=3*29. Only D fits. You don't even need to solve anything.

Answer: D
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It is observed that 2/7 f the balls that have red color also have 02 Sep 2019, 19:21
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KSBGC
There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f the balls that have red color also have green color while 3/7 of the balls that have green color also have red color. What fraction of the balls in the jar has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT

We can create the equations:

r + g - both = 87

and

(2/7)r = (3/7)g (note: either side of the equation = both)

2r = 3g

r = 3g/2

Substituting, we have:

3g/2 + g - 3g/7 = 87

5g/2 - 3g/7 = 87

Multiplying by 14, we have:

35g - 6g = 87 x 14

29g = 87 x 14

g = 3 x 14 = 42

Therefore, both = 3(42)/7 = 3(6) = 18, and balls of both colors are 18/87 = 6/29 of all the balls.

Answer: D
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KSBGC
There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f the balls that have red color also have green color while 3/7 of the balls that have green color also have red color. What fraction of the balls in the jar has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT

r+g-both+none=87…r+g-both=87
2/7*r=3/7*g…r=3/2*g
3/2*g+g-3/7*g=87…5g/2-3g/7=87
35g-6g/14=87…g=87*14/29
both/total=(3/7*g)/87=(3/7*87*14/29)/87=6/29

Ans (D)
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It is observed that 2/7 f the balls that have red color also have 12 Nov 2020, 01:11
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R:(R&G) =5:2
G:(R&G) =4:3
As red and green balls are equal we need to make their respective proportions commensurate. By multiplying first proportion with 3 and 2nd with 2 we get,

R:(R&G)=15:6
G:(R&G)=8:6

so R:G:(R&G)=15:8:6

the required fraction = 6/(15+8+6)=6/29

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hi
i am not able to understand logic behind taking red ball as 7/2 and green ball as 7/3 in this question. could anyone can explain
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It is observed that 2/7 f the balls that have red color also have 02 Dec 2020, 09:46
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I didn't solve the problem.

Since the balls count should be a positive integer, I checked the options whose denominator exactly divides 87.

Only Option D has 29 which exactly divides 87.
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